# SO(2,1) - Haar measure, exponential map

1. May 5, 2013

### Bobhawke

I wasn't quite sure where to put this, so here goes:

I am trying to find out some facts about the group SO(2,1). Specifically; Is the exponential map onto? If so, can the Haar measure be written in terms of the Lebesgue integral over a suitable subset of the Lie algebra? What is that subset?

If the exponential map isn't onto, then for what subset of the Lie algebra is it at least 1-1 (injective)?

I was hoping there would be a reference out there that explained these things, but have so far been unable to find it.

Thank you.

2. May 5, 2013

### fzero

SO(2,1) is not connected. There is a component that is connected to the identity, called $SO^+(2,1)$, for which the exponential map is surjective. This is assumed in all of the physics literature on the Lorentz group, but I am at a loss to come up with a simple convincing proof at the moment. The surjectivity is claimed as a result of Nishikawa in prop 1.6 in http://www.heldermann-verlag.de/jlt/jlt07/DOKHOFPL.PDF.

Since the exponential map is surjective, you should be able to use the Maurer-Cartan 1-form $\omega= g^{-1} dg$ to build the Haar measure using the appropriate angle and real parameters.

3. May 6, 2013

### Bobhawke

Could you say something like: Each element of the Lie algebra generates a 1-parameter subgroup in the connected part of the group, and locally the basis elements of the Lie algebra take you in orthogonal directions. Therefore you can string together a series of curves which take you from the origin to any point which it is path connected?

4. May 6, 2013

### fzero

What you say is true for compact groups. It may also be true for the connected components of matrix groups. I'm not familiar enough with the esoteric details to say with 100% confidence. The counter-examples to surjectivity for matrix groups that I've found in the literature all seem to be related to disconnectedness. On the other hand I know that for diffeomorphism groups, the exponential map is not generally surjective, and this is directly related to noncompactness, rather than disconnectedness.

For SO(2,1) we have some extra tricks available, since it is isomorphic to $SL(2,\mathbb{R})$. The connected component is $PSL(2,\mathbb{R})=SL(2,\mathbb{R})/\{\pm I\}$. So we can use the fact that the geometry is that of the unit tangent bundle of the hyperbolic (upper-half) plane (see http://en.wikipedia.org/wiki/SL2(R)#Topology_and_universal_cover).